A good isosceles triangle problem

Thanks to Dhrubajyoti for the suggestion! A similar problem was asked to class 10 students in India (Q33 from SMO junior 2016). I admit that this one stumped me. Can you figure it out?
Solution presented: post by Thomas Hill on Quora
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Other solutions
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[@gman3243]

His geometry puzles are always creating other shapes.

[@abhinavbhutada9b484]

😂 Apply sine rule in the two triangles to easily chase theta. There is no construction required for thos who know olympiad mathematics.

[@aryanpatel453]

Ugot A stack of likes

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@aayushpaswan2941]

@@abhinavbhutada9b484 its not right sngled triangle to apply trignometry 😑😑

[@jokarmaths7771]

amazing .............kzbin.info/www/bejne/fai3p5qljpilo8U

[@steveyoth1]

I dropped a vertical line from C and extended the line AB to the left to form a new triangle. Let the equal sides be length 1. The new triangle has sides of sin(10deg) and cos(10deg), which is .9848 and .1736. The length of CB is then SQRT(.9848^2+1.1736^2) = 1.532. Then the angle in question is just INVTAN(.9848/(1.532+.1736)) which is 30 degrees.

[@narada6329]

I also did that, nice

[@donasaloum9154]

We can solve any geometric problem using trigonometric ratios But skill and intelligence is in solving it by drawing or by simple geometric rules

[@sym959]

@@donasaloum9154 I agree. But using trigonometry is sometimes the easiest (and you could argue that its smart to use the most straightforward method) and most convenient method.

[@donasaloum9154]

@@sym959 😌😊

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@wesleysuen4140]

Yet another approach: Set up the figure in the Cartesian plane with A at (0,0) and B at (1,0). Then C is at (cos100°,sin100°) =(-cos80°,sin80°), and D at (d,0) where d=sqrt((1+cos80°)²+sin²80°) =sqrt(2(1+cos80°))=2cos40°. tan∠ADC = -(slope of DC) = sin80°/(2cos40°+cos80°) = cos10°/[cos40°+(cos80°+cos40°)] = cos10°/[cos40°+2cos60°cos20°] = cos10°/(cos40°+cos20°) = cos10°/(2cos30°cos10°) = 1/sqrt(3) so ∠ADC = 30°.

[@ravifaujdar3178]

My favorite Cartesian method.....nice solution

[@animezoneamv9116]

Man I completely forgot distance formula. It took like 5 mins to understand you used that. If I can meet you physically anywhere, I wouldn't refuse to award you with money for this truly spectacular solution.

[@hoclyonline2069]

I have another method to solve it which is quite interesting draw angle BCT = 10 degrees, draw square AH with CT, Let N be the intersection of AH and BC, Draw AT1 perpendicular to BC. Then I will prove that AH = CT1 = half BC = half AT => AT = AD or T will coincide with D. So angle ATC = angle ABC - angle BCT (outside angle of triangle) = 40 degrees - 10 degrees = 30 degrees

[@strictlyeducationalmagick]

I usually can't understand what you're talking about, but was able to spit this one out in 15 seconds,

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@romain.guillaume]

Actually I used a more « brutal » method. I created a right triangle from the height of C and then compute the coordinates of the points C (x=0, y=AC*sin(180-100)) and D (OD=OA+BC => OD= AC*cos(80)+AC+Al-Kashi in ABC). Then in the triangle ODC can now use the tangent of angle D and all the AC cancelled each other.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@xz1891]

Actually A' is the center of the inscribed circle for triangle ABD, ez to figure out all angles via circumferential angles.

[@Aiden-xn6wo]

ABD is a straight line.

[@vhm0814]

@@Aiden-xn6wo He means ACD

[@xz1891]

@@vhm0814 yes, ACD, thank you

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@jokarmaths7771]

amazing .............kzbin.info/www/bejne/fai3p5qljpilo8U

[@kienthanhle6230]

My solutions: Let AB = AC = a CB = b CD = c By using the law of cosine (this is pretty easy to prove, if you don't know what it is, just look it up), we have: b^2 = a^2 + a^2 - 2.a.a.cos 100 c^2 = a^2 + b^2 - 2.a.b.cos 100 (AC^2 =) a^2 = b^2 + c^2 - 2.b.c.cos BDC 2.b.c.cos BDC = b^2 + c^2 - a^2 cos BDC = (b^2 + c^2 - a^2) / (2.b.c) = sqrt(3)/2 (2.b.c is always greater than zero so yeah) Therefore BDC is equal to 30 This doesn't involve drawing any extra line, but it will be kinda hard to do without a calculator. So if I was in a contest where calculators were not allowed, this method would be less than ideal.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@vladimirrainish841]

There is a straightforward trigonometrical solution arctag(sin80°/(cos80°+2cos40°) cos40 (and sin/cos 80) is a computable value (via cos20°) since 4cos³20° - 3cos 20° = cos60 = ½

[@JoshuaCassel]

I love and hate your show some. It has humbled and irritated me but I keep watching because I love these puzzles.

[@vishweshpatil8451]

Can also be solved using Sine Rule in triangles ACD and BCD.

[@jokarmaths7771]

amazing .............kzbin.info/www/bejne/fai3p5qljpilo8U

[@dojo2836]

@@hafizrahmanelikkottil2107 Hello, how do you go from sin140/sin100 = sinθ/sin(80-θ) to sin40/sin80 = sinθ/sin(80-θ) ? Thank you

[@jokarmaths7771]

amazing ...kzbin.info/www/bejne/jWTFh3WDe5efedE

[@dojo2836]

@@hafizrahmanelikkottil2107 oh yes of course! nice proof :)

[@HeckaS]

@1:26 Why do you assume AD is equal to BC? This isn't indicated anywhere.

[@devondevon4366]

did something differently using law of cosines to get 30 degrees let assume the AB=AC=2, then using the law of cosine BC= 3.06, hence AD=3.06 since both are equal. Hence BD= 1.06. Since angle CBD= 140 degree, and line BC= 3.06 and line BD= 1.06, then using law of cosine angle D=30 degrees answer

[@jokarmaths7771]

amazing ...............kzbin.info/www/bejne/fai3p5qljpilo8U

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@Rolexor]

Presh, I enjoy your videos. I didn’t see this solution coming from a mile away. Having stated that, let me further state, your videos have a much higher difficulty level after I’ve had a couple of drinks. Maybe you should do some videos that explain the solution for inebriated people.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@sundareshvenugopal6575]

If two sides of an isosceles triangle are equal, the angles opposite those two sides are also equal.

[@jokarmaths7771]

amazing ...kzbin.info/www/bejne/jWTFh3WDe5efedE

[@srividhyamoorthy761]

thank you for a awesome solution which is fascinating i am also indian and i am 8 th grade

[@liuben3384]

there is a flaw that when you make the point A’, how can you assure it is outside of line CD? This should be proved first

[@liuben3384]

it is must though it is easy

[@quandaledinglegaming9930]

i was doing the problem without even noticing angle 100💀

[@242math]

very well done, thanks for sharing

[@jokarmaths7771]

amazing .......kzbin.info/www/bejne/fai3p5qljpilo8U

[@jokarmaths7771]

amazing .............kzbin.info/www/bejne/fai3p5qljpilo8U

[@jokarmaths7771]

amazing ...kzbin.info/www/bejne/jWTFh3WDe5efedE

[@phungcanhngo]

That's amazing!

[@AkashRaj-nq7qm]

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[@jokarmaths7771]

amazing ...............kzbin.info/www/bejne/fai3p5qljpilo8U

[@kneD-e]

Bruh this is in NCERT class 9 or 10

[@angelinageorge2278]

Exactly this is in 10th Maine 9th ncert Pura kar liya

[@darmandixus6318]

I got that angle BDC=atan(sin80*sin140/(cos80*sin140+sin100))

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@jokarmaths7771]

amazing .......kzbin.info/www/bejne/jWTFh3WDe5efedE

[@Awesome-ct7vr]

I tried to solve this for like an hour. Saying no way, thats impossible. Its lacking info. Only to realize that i didnt copy the part where AD = BC

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@jokerdady8470]

Sir 1^1 = 1----i and 1^0 = 1 ------ii equating i and ii we get 1=0 sir where is an error plzz help me to find error sir plzzz make video on this topic

[@2012tulio]

using trigonometry the angle is 30

[@adamdosa]

🌹

[@양휘찬-x2x]

Very easy

[@balak2212]

I DIIID IT WIT MA OWN METHOD HAHAHAHHAHA AND MIND YOU I AM IN CLASS 10 WHO HOOOOOOO( THIS IS MY FIRST TIME SOLVING A QUESTION ON THIS CHANNELLL I AM HAPPPYYYYYYYYYYY)

[@g_68sreemanpattanaik63]

Sir Please Also try Jee Advanced exam Math Problems

[@jokarmaths7771]

amazing ...............kzbin.info/www/bejne/fai3p5qljpilo8U

[@AkashRaj-nq7qm]

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[@lameuerte]

wow

[@jokarmaths7771]

amazing ...........kzbin.info/www/bejne/fai3p5qljpilo8U

[@peterevans3310]

I found 30° using cosine rule

[@WaiWai-qv4wv]

💯💯💯💯💯💯💯❤️❤️❤️❤️❤️

[@jokarmaths7771]

amazing .......kzbin.info/www/bejne/jWTFh3WDe5efedE

[@10names55]

Ez for ioqm aspirants

[@parikshitkakoti7289]

Problem suggest কৰাজন অসমীয়া ল'ৰা হ'ব...নামটো তেনে লাগিছে

[@jokarmaths7771]

amazing .......kzbin.info/www/bejne/fai3p5qljpilo8U

[@meomymieu9982]

Can anyone help me out with this problem? 0≤a,b,c≤1 a+b+c=2 P=ab/(1+ab) + bc/(1+bc) +ac/(1+ac) Min, Max P=?

[@xz1891]

I discussed a few cases, all base on "AM-GM inequality", Case 1, either =0, Case 2, 1 of a b c =1 Case 3, general Then summarize all cases, my answer is, when a=b=c =2/3, got max =12/13, min is possibly =1/2,

[@jokarmaths7771]

amazing ...........kzbin.info/www/bejne/fai3p5qljpilo8U

[@IEEMAZ_Convoluted_14.2.8.5]

Changed it to background black eh?

[@jokarmaths7771]

amazing ...........kzbin.info/www/bejne/fai3p5qljpilo8U

[@keminoff]

Less Geometry, more Algebra!

[@IEEMAZ_Convoluted_14.2.8.5]

You want that or this is more algebra

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@sergiobaldares5013]

Point A’ is actually the midpoint between A and the intersection of the A-Symmedian with the circumcircle of ABC. This is a common configuration for high end mathematical Olympiads like the IMO

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@leif1075]

What circle? Presh didn't use or draw a circle?

[@obelusillumina6077]

@@leif1075 Circumcircle is a circle that contains the vertices of a triangle on its circumference. Basically Sergio is referring to the circle that you get when you try to construct one containing points A,B,C. (I'm assuming this is what you're asking about. If not then it's fine.)

[@vongocvinh9734]

Even though I'm not good at English, I still understand what you mean because you illustrate the picture so well😄

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@WestExplainsBest]

His visuals are top notch and that is another benefit to Math - it's a universal language!

[@mrcastor9514]

How do you comment in english then.

[@codewithsuw9014]

translate maybe

[@rdblocks5490]

@@mrcastor9514 he said hes not good not that he doesn't know it at all

[@omrquliyev1905]

Another solution: construct equilateral triangle EBC on side BC, then angle EBA=ECA =100°, since angles CBE=BEC=ECB =60°, which means that ∆EBA=∆ECA=∆ADC - 2 sides and angle between them: EC=EB=AD, AC=AB, angles EBA=ECA=CAD=100°, it follows that angles AEB=AEC=CDA, but AEB+AEC=60°, thus each AEB=AEC=CDA=30°, that is required to prove.

[@ahmedMohamed-np1vj]

I respect this channel very much, and I love its explanation. But this time I see that explanation in comments is better than that in the video.

[@gurpreet_singh6894]

@@ahmedMohamed-np1vj yeah

[@ujwalbiradar325]

Damm bro!!! Extraordinary thought process...!!love u ! Where r u from? Pls reply

[@kienthanhle6230]

Damn, that's so cool bro. How did you come up with that?

[@jokarmaths7771]

amazing .............kzbin.info/www/bejne/fai3p5qljpilo8U

[@patel_pankaj_69]

Also get this ∠A=100° and ABC is isosceles triangle thus ∠B is 40. Now, Suppose E is the angle near B. Thus, ∠B and ∠E is the angles of a linear pair. So, ∠E= 180-∠B= 180-40= 140° Then in Question, given AD=BC AD= 3BD So BC:BD=3:1 Opposite angles are also 3:1 thus,D is 3 times ∠C ∠D=3∠C........(eq.1) Now ∆EDC, Sum of all angels are 180° ∠E+∠D+∠C= 180° But ∠E is 140° (got earlier) 140+3∠C+∠C=180° 4∠C=40° ∠C=10° Now from (eq.1) ∠D=3∠C=3*10=30° Thus, ∠D=30° ✌️💫❤️

[@Nxck2440]

Solution with no drawing extra lines: Let AC = AB = 1 BC = sqrt(1^2 + 1^2 - 2(1)(1) cos 100) [cosine rule in triangle ABC] BC = sqrt(2 - 2 cos 100) BC = 2 sin 50 [half angle formula] AD = BC = 2 sin 50 CD = sqrt(1^2 + 4 sin^2 50 - 4 sin 50 cos 100) [cosine rule in triangle ACD] CD = sqrt(1 + 4 sin 50 (sin 50 - cos 100)) CD = sqrt(1 + 4 sin 50 (sin 50 + sin 10)) [reflection identity] CD = sqrt(1 + 8 sin 50 sin 30 cos 20) [sum to product identity] CD = sqrt(1 + 4 sin 50 cos 20) CD = sqrt(1 + 2 sin 70 + 2 sin 30) [product to sum identity] CD = sqrt(2 + 2 cos 20) CD = 2 cos 10 [half angle formula] ABC is isosceles -> angle ABC = angle ACB -> angle ABC = (180 - 100)/2 = 40 -> angle DBC = 140 sin BDC = (2 sin 50 / 2 cos 10) * sin 140 [sine rule in BDC] sin BDC = sin 50 sin 40 / cos 10 sin BDC = 1/2 * (cos 10 - cos 90) / cos 10 [product to sum identity] sin BDC = 1/2 --> BDC = 30 degrees

[@bkj556]

Kudos to you 🙌

[@jokarmaths7771]

amazing .............kzbin.info/www/bejne/fai3p5qljpilo8U

[@ramongutierrez6295]

Ditto! I just got done writing down the same solution :)

[@Tiqerboy]

True, didn't say you couldn't use Trig. Presh should have stated "without using Trigonometry, solve using only elementary methods" if he wanted his solution to be the one they wanted.

[@timeisapathwalkingtounderstand]

Huh?

[@hoclyonline2069]

dịch sang Tiếng Việ nha kẻ góc BCT =10 độ , Kẻ AH vuông với CT, Gọi N là giao điểm của AH và BC, Kẻ AT1 vuông góc với BC. Sau đó tôi sẽ chứng minh đc AH = CT1 = một nửa BC = một nửa AT => AT =AD hay T sẽ trùng với D. Vì Vậy góc ATC = góc ABC - góc BCT ( góc ngoài tam giác) = 40 độ - 10 độ = 30 độ

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@charlesbromberick4247]

This problem can b solved rather easily with law of cosines and sines, and without pulling strange constructions from who knows where.

[@thetruthudontwannahear740]

Yeah but with sin n cos that didnt turn out to be the exact answer it was somethinh like 32.7°🤷🏽‍♀️ and now im confused

[@ahmadalghamdi9002]

Try to avoid drawing many figures

[@anandk9220]

I used sine rule in triangles ABC and ADC to express AD and BC in terms of AC. Then used online calculator to get ADC = 30°.

[@helloworld7448]

Same bro

[@MotassinShehab]

how can you use sine rule? the angle is not 90 degrees

[@2012tulio]

@@MotassinShehab in triangle BDC use first the cos rule to get a relation between CD , BD and BC and you have the angle DBC = 140 , then use the sin rule to find the angle .

[@animezoneamv9116]

@@2012tulio What's the sine rule and cosine rule ?

[@lox7182]

Will you be allowed to use a calculator in an olympiad setting?

[@TrumpetGuy360]

I solved this using cosine rule. BC = sqrt(AB^2+AC^2-2*AB*AC*cos(100)) I just subbed in 1 for side AB and AC to get BC = 1.5321 We're rounding, but I assumed that the final solution would be close enough to a whole number and it makes the math easier on me. Then we can find side CD using cosine formula as well. CD = sqrt((1.5321-1)^2+(1.5321)^2-2*(1.5321-1)*1.5321cos(140)) = 1.9696 Then reverse cosine rule to find angle D. cos(D)=(1.9696^2+(1.5321-1)^2-1.5321^2)/(2*1.9696*(1.5321-1)) = 0.86587 D = 0.523701 + 6.28319*n radians 0.523701 rad = 30.006 degrees So I guess the answer is supposed to be 30 degrees if I didn't round.

[@xz1891]

Most cumbersome way

[@TrumpetGuy360]

@@xz1891 You can tell I’m an engineer and not a mathematician

[@xz1891]

@@TrumpetGuy360 any non special angle s trig calculation is to be avoided in a math test 😁

[@Aiden-xn6wo]

@@TrumpetGuy360 I think I can tell that.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@peterkwan1448]

I solved it this way: Extend DA in the direction towards A to a point H such that AHC is a right triangle. Since ∠BAC is 100°, ∠HAC=80°. Suppose CH=1. Then AH=cot 80°=cos 80°/sin 80°. Since AB=AC, ∠ABC=40°. Then BC=CH/sin 40°=2 cos 40°/sin 80° (by double angle identity for sine). Since BC=AD, HD=(cos 80°+2 cos 40°)/sin 80°. By sum and difference formulae, cos 80°+2 cos 40°=cos 60°cos20°-sin 60°sin20°+2cos 60°cos20°+2sin 60°sin20°=3cos 60°cos20°+sin 60°sin20°=√3/2(√3 cos 20°+sin 20°). Similarly, sin 80°=sin 60°cos 20°+cos 60°sin 20°=1/2(√3 cos 20°+sin 20°). So HD=√3, and CHD is a 30-60-90 triangle, with ∠CDH=30°.

[@ramongutierrez6295]

Another way is if you set the two equal sides to 1. Then it is just a law of sines and law of cosines problem. There is no additional drawing or reconfiguration needed. Good stuff. I love it when different avenues lead to the same solution :) LoS BC/sin(100)=1/sin(40) BC=sin(100)/sin(40)=1.532 AD=1+BD=BC, so BD=BC-1 BD=1.532-1=0.532 LoC DC^2=BC^2+BD^2-2BC*BDcos(140) DC=1.97 LoS DC/sin(140)=BC/sin(BDC) DCsin(BDC)=BCsin(140) BDC=sin^-1(BCsin(140)/DC) =sin^-1(0.5)=pi/6 =30 deg

[@bishnupal8649]

Great 👍👏

[@sswy1984]

this kind of questions, you should always find an equilateral triangle somewhere. make a point E below line AB, such ABE is equilateral triangle, then connect CE. this is another solution, similar.

[@viralvideo2036]

Who knows that Angle BCD+Angle CDB =Angle CBA ? i think theorem name is Distal Angle theorem

[@HassanLakiss]

Thank you for a very interesting and challenging question. Your approach was very interesting and is a fine example of thinking out of the box. Your explanation as usual is very clear and easy to follow and understand. Thank you Sir.

[@rohangeorge712]

i used trignometry, the non creative solution

[@HassanLakiss]

@@rohangeorge712 Using trigonometry is fine.

[@WestExplainsBest]

Agreed! I would have never thought of that solution method.

[@jokarmaths7771]

amazing .............kzbin.info/www/bejne/fai3p5qljpilo8U

[@gman3243]

His geometry puzles are always creating other shapes.

[@jokarmaths7771]

amazing .............kzbin.info/www/bejne/fai3p5qljpilo8U

[@ANDROIDPOSTMORTEM]

I can't stop appreciating your solution ❤... Anyway, I solved it using trigonometry within some minutes

[@charle-edouardsavoie6634]

It's the proper way to do it hahaha

[@ANDROIDPOSTMORTEM]

@@charle-edouardsavoie6634 Proper way to do???... Which proper way?

[@salkabalani1482]

Thank you so much for the black background. The old videos with white background were searing into my retina, but had to watch them because of the fantastic content. Keep up the good work

[@shri4tunes]

How does one logically come up with these constructions? Looked at the solution and I admit I would never have thought about it.

[@Skandalos]

Here is the lazy solution: let AB=1 and angle ADC be alpha BC = 2 cos(40°). By sine rule for the triangle ACD: 2cos(40°)/sin(80-alpha) = 1/sin(alpha) By some sine identity: sin(80°)cos(alpha) - cos(80°)sin(alpha) = sin(alpha)*2cos(40°) Dividing by sin(alpha): sin(80°)cot(alpha) - cos(80°) = 2cos(40°) Solving for cot(alpha) = (2cos(40°)+cos(80°)) / sin(80°) calculator says: alpha = 30°

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@abhishekpatil1063]

Let AB=AC=1 unit. So using cosine law for triangle ABC, Cos100=(1+1-BC^2)/2 So BC^2=2.3473 So BC=1.532. But BC=AD since it is given data. So AD=1.532. ButAB=1, So remaining BD=1.532-1= 0.532. Now using Sine law in triangle BCD, Let AngleBDC=x (say). So sine x/1.532 =sine(40-x)/0.532 So solving further x=30 degrees.

[@jokarmaths7771]

amazing ...............kzbin.info/www/bejne/fai3p5qljpilo8U

[@Everyone.needJustice]

Tringle BCD =180° Then,.

[@2222ut]

I love these problems. Thanks. I tried to think of a solution using Mathematica. I used a SAS triangle: SASTriangle[1, 100*Degree , 1] To return 3 points. The Hyp was: Sqrt[2*(1 + Sin[Pi/18])] The new larger Triangle is now: SASTriangle[1, (5*Pi)/9, Sqrt[2*(1 + Sin[Pi/18])]] The first point is at {0,0}, and the third point given is {x,y}, so the angle would be ArcTan[x,y]*180/Pi I could not simplify the equation, but numerically it came out to 30 Deg (180./Pi)*ArcTan[(Cos[10*Degree]*Sqrt[2*(1 + Sin[10*Degree])])/ (2 + Sin[10*Degree]*(2 + Sqrt[2*(1 + Sin[10*Degree])]))] 30.00000000000000 [Added:] Oh. I got it. If you expand the functions: ArcTan[(Cos[10*Degree]*Sqrt[2*(1 + Sin[10*Degree])])/ (2 + Sin[10*Degree]*(2 + Sqrt[2*(1 + Sin[10*Degree])]))] It converts to: ArcCot[(Sec[Pi/18]*(2 + Sin[Pi/18]* (2 + Sqrt[2*(1 + Sin[Pi/18])])))/ Sqrt[2*(1 + Sin[Pi/18])]] Now, it reduces: FullSimplify[%] Pi/6 Great problem! Thanks.

[@Ruleor]

i’m 16 do i need to learn this

[@hyper_phantom_darxteel]

I like to brute force through questions so I just used the cosine formula and the sine formula

[@shadrana1]

Let CA=AB= 1 unit angle ACB = angle ABC = 40deg. sin(100)/CB=sin CDA/1 CB=2 cos40 =AD=2 cos40 Extend DA to the left and drop a perpendicular to intercept DA extended at A' CA'=sin80=2sin40*cos40 AA'=cos80 A'D=A'A+AD=(cos80+2*cos40) tan(ADC)= CA'/A'D=(2sin40*cos40)/(cos80+2cos40)=1/sqrt3 arctan(1/sqrt3)=30 deg. and that is our answer.

[@viswanaths_travel_jam]

SSC CGL walo ko meraa salaam

[@Chrisoikmath_]

Let me give a message from here: Please, stop the war! ☮

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@quocbao1297]

I have different solution, let's BAX is a equilateral triangle, XBC = 100°, so triangle XBC = triangle CAD

[@hishivam004]

My calculation shows 26.67 degree, but my common sense shows 30 degrees

[@wesleysuen4140]

I reflect the figure wrt the perpendicular bisector of BD and symmetrize it into an isosceles trapezium. Let’s denote the images under the reflection of the marked points by the same letter with a prime(‘). Join CA’ and CC’. In particular, with B’=D, D’=B, note that i. CC’ // AA’, and ii. A’BC is an isosceles triangle (since BA’ = D’A’ = D’B’+B’A’ = BD+AB = AD = BC). From i., ∠BCC’ = ∠ABC = 40° and ∠DCC’ = ∠ADC. From ii., ∠BCA’ = ∠BA’C = ∠ABC /2 = 20°. Now, ∠BCD can be written in two ways: ∠ABC-∠ADC = ∠DCC’-∠BCC’ Then 40°-∠ADC =∠ADC-20° and so ∠ADC=30°.

[@bkj556]

Kudos 🙌🤕

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@jokarmaths7771]

amazing ...............kzbin.info/www/bejne/fai3p5qljpilo8U

[@Grimmet2137]

I was like "oh this one is 40*, and this one is a bit smaller, I guess it's 30*"

[@davidcahan]

I see now why you couldn't solve. What math level would this be considered appropriate for? Certainly not high school or even undergrad college right?

[@洪米堂]

i took a heavy calculator to bombing this problem then i got the answer=30deg

[@Mathematician6124]

See as fun tends to infinity's geo puzz 69, there you will find many solutions including mine

[@אלוןסלעי]

Give AC=5 for instance and solve it with simple high school methods.

[@kurzackd]

pretty sure that's among the more complicated ways to solve this. Theoretically there's easier ones.

[@vondahartsock-oneil3343]

Sorry, it's me again. The one who does it in my head. Looking at it I say 30 degrees.

[@mibsaamahmed]

That was a neat solution but if you can then please solve these types of questions with other methods such as the law of sine

[@hishivam004]

Any other methods.. Like ration of sides and angles.

[@hoangnhan6612]

I need to watch this vid 2 times to relise this is just a 6 grade problem

[@DLatryShankLi]

well i got lost when he said BDA' is a triangle shudent it be ADA'?

[@bobgade6733]

Is c-a and a-b are congruent, the angle would not be 100

[@fernandoalmer3312]

Anyone can help me with this problems? √(4+√(16+√(64+√...)))) *Continue to infinitely many

[@jokarmaths7771]

amazing ...........kzbin.info/www/bejne/fai3p5qljpilo8U

[@zencasaw2998]

I need this video for my math exams and homework

[@sutapadey5274]

another method construct equ. triangle ADE so that E is above AD so CAE = 100-60 = 40 notice that: •

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@jokarmaths7771]

amazing ...............kzbin.info/www/bejne/fai3p5qljpilo8U

[@2w2w3e]

after You know the ACA` is the regular triangle You know that

[@ФилиппЛыков-д8е]

I added two copies of the initial triangle rotated 100° and 200° around point A so that the point C of the first triangle becomes point E of the second and point F of the third one. Then, in the third triangle (AEF) I constructed EG the same way as CD in the first triangle, so that F is between A and G. Then I constructed GD. The angle DAG is 60° since DAC+CAE+EAG=300°. Hence GD = AD = AG. But AD = BC and EC = BC so that GD = EC and CEGD is a rectangle. Hence angle GDC = 90° and BDC = GDC - GDA = 90° - 60° = 30 °.

[@NeverTalkToCops1]

Riddle me this: A woman with one blue eye and one green eye has an 8 year old bobcat and an 89 year old father. How old is the woman?

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@davidosborne1306]

I just used the law of sines once to find the length of BC, (using 1 as the lengths of AB and AC, and knowing all three angles), then law of cosines with triangle BCD to find the length of CD (with two known side lengths and one angle), then with all side lengths and one angle (of either triangle ACD or triangle BCD) known, just use the law of sines to find the angle BDC (or ADC, works either way, in fact I did it both ways to double check myself). I also just did the whole thing again, while I was editing this post. I thought I had an idea that might eliminate 1 step, and maybe solve it without using the law of cosines, but I ended up having to use exactly the same three steps that I used the first time.

[@vondahartsock-oneil3343]

Hell I just looked at it and figured it in my head in like 10 seconds lol. I never do "math" the "academics way" but I won awards in college for it.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@utubewatcher1344]

or you could use sine and cosine rule

[@jokarmaths7771]

amazing .............kzbin.info/www/bejne/fai3p5qljpilo8U

[@academicunboxing1402]

Just out of curiosity, how tall are you?

[@Tiqerboy]

Solved it more or less the way you did. To me it was obvious to transpose the isosceles triangle as you did to start the problem since it was given that AD = BC. Oh, you might want to explain, for your diagram with the transposed Isosceles triangle to make sense (i.e. point A' lies outside triangle ADB and not inside), that angle ABC > angle ADC. I know it's obvious but you should explain why. You may not be aware but on my PC use of KZbin, when you mouse-over a video, KZbin begins to play it through and when it does, it gives this problem away, because the crucial hint is given. Just so you know.

[@pflintaryan8336]

sin(40°) = tan(angle(BDC))

[@vanguelder]

witch tool you use to make this videos?

[@alexcwagner]

I just cranked it out numerically and got 30 degrees. When I went back and tried it analytically, I ended up with sin(theta)/cos(theta+10) = sin(30)/sin(40) and it's obvious that theta=30 satisfies the equation, but it seems like kind of a silly way to get there.

[@giuseppemalaguti435]

sinx=sin100sqrt3, ho sbagliato!!! Ho usato il teorema dei seni per i 2 triangoli, il teorema di carnot.... Comunque risulta sinx=0,5....x=30

[@paulgets2737]

I didn't see this interesting constructione and i solved the problem using first cosine law and then the sine law.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k

[@spacemario]

Wait so to solve this I'm dependant on the graphic? edit: no. Since we know x is smaller than 40°, we already know the vertex A' will be above the segment CD

[@maihuphoenix3301]

Same ,i thought the exact answer

[@pki0273]

reaLol

[@jokarmaths7771]

amazing ...........kzbin.info/www/bejne/fai3p5qljpilo8U

[@Mark-cz2qe]

arctan（sin80sin40/cos80+sin100）

[@whitegalactico7152]

My ans was 20°.I thought I was right but after seeing the solution my thought was wrong!Anyway the solution helps me think in different ways and it also increases my intuition

[@HackedPC]

This is the first problem on this channel that I couldn't solve.

[@AkashRaj-nq7qm]

kzbin.info/www/bejne/fmjCaGSIYsiUb9k